# Can a syringe be used as a hanging scale if needed?

FYI: This is not a homework q; I am just curious... You can assume that...

• the syringe tip cap creates a "perfect" seal, i.e., keeps air from flowing in/out!
• the syringe is completely depressed before starting the experiment.
• the outside barrel has metric ruler.
• we have access to accurate temperature & atmospheric pressure readings.
• the syringe is hung in the following way...

Is the following equation correct for determining the mass of the weight, based on how far it causes the plunger to depress...

$$m = \frac{A\cdot B\cdot x}{a\cdot L}$$

where...

• A = Surface area of the (black) stopper face.
• B = Bulk modulus of air given temperature & atmospheric pressure readings.
• x = Plunger displacement (due to suspending weight) after waiting for a "long" time.
• a = Acceleration due to gravity.
• L = Length of syringe barrel.

## 1 Answer

Yes you are basically right, and I will add a historical note. When Robert Boyle (1627-1691) did his experiments which led to what we now call Boyle's law, he compared air to a spring and wrote of "the spring of the air", very much in the same spirit as you are doing here. He titled his book "New Experiments Physico-Mechanicall, Touching the Spring of the Air, and its Effects"; it was an important contribution to the development of experimental physics. His approach was along similar lines to yours---thinking through in precise mechanical terms the consequence of a change in volume of a given amount of air. There is one slight drawback in the use of an ordinary syringe, in that they normally have a lot of friction. The experiment would work better with a metal piston and cylinder, with o-rings and grease a bit like those used in a car engine.

• Yes, I thought about the friction... The syringe manufacturers could VERY accurately measure the force ($= F_{\text{tc off}}$) required to overcome the friction between stopper & barrel (i.e., w/ the tip cap off) engraving the outcome; then my equation would look like this... $$0 = F_{\text{tc on}} + F_m \Longrightarrow 0 = \left(\frac{A\cdot B\cdot x}{L} + F_{\text{tc off}}\right) + \left(-m\cdot a\right) \Longrightarrow m = \frac{A\cdot B\cdot x}{a\cdot L} + \frac{F_{\text{tc off}}}{a}$$ Does it still look correct, after I've made the update? Jun 2, 2019 at 23:42